Question

The slope of the graph drawn between $\ln k$ and $\frac{1}{T}$ as per Arrhenius equation gives the value $\left(R=\right.$ gas constant, $E_{a}=$ Activation energy)

• A. $\frac{R}{E_{a}}$
• B. $\frac{E_{a}}{R}$
• C. $\frac{-E_{a}}{R}$
• D. $\frac{-R}{E_{a}}$

Answer 1

Answer: (C)

We know that, Arrhenius equation,

$=A e^{-E_{a} / R T}$



On taking log on both sides, we get

$\log\, k=\log A-\frac{E_{a}}{2.303 \,R T}$

Or, $\ln \,k=\ln A-\frac{E_{a}}{R T}$

$\therefore $ A graph between $\ln \,k$ and $1 / T$ is a straight line (according to the equation of straight line, viz. $ y=m x+c)$ with $-\frac{E_{a}}{R}$ slope (negative) and In A intercept.